# Is Euler characteristic of a binary image truly a topological invariant?

I was reading the paper: "A fast algorithm for computing the Euler number of an image and its VLSI implementation (Dey et al., 2000)". On the first page they say:

Topological properties serve the purpose of representing the geometric shape of an image. They remain invariant under any arbitrary rubber-sheet transformation and hence are very useful in image characterization for matching shapes, recognizing objects, image database retrieval and many other numerous image processing and computer vision applications. An important topological feature of an image is the Euler number (or genus) which is the difference between the number of connected components and number of holes.

Now, let us suppose we represent our binary image as consisting of many tiny square pixels. The white clusters are our "connected components" and the black clusters are our "holes".

According to the definition given in the paper, the $\chi$ (Euler number) for both the figures is $\text{Number of Connected Components - Number of Holes} = 1 - 1 = 0$. However, to me, the first figure does not in any way seem like a rubber-sheet transformation or homeomorphism of the second figure! But then, how is $\chi$ a topological invariant? Am I missing something?

• Euler number being topological invariant means two homeomorphic objects get the same euler number, doesn't mean that two objects with same euler number are homeomorphic.
– user43326
Aug 4, 2018 at 17:05
• This question would fit better at math.stackexchange. Aug 4, 2018 at 18:18

Homeomorphism (the rubber-sheet transformations - or, more properly, homeomorphism class) is a very strong kind of invariant. But there are stronger kinds (e.g. equality, diffeomorphism class, Lipschitz equivalence) and weaker kinds (e.g. homotopy class, $\pi_1$, cardinality).